12; then we have 1 x 4 = 1.587; 4 x 8 = 3.174; and 3/9 × 12 = 4.762, which are in the linear ratio of the sections, or 1, 2, 3. Similarly with the cast-iron girders D, E, F, in Fig. 153, we have 1 x 5 = 1·71; 3/4 x 10 = 3.42; and 9 × 15 = 5.13; which are still in the linear ratio of the sections, or 1, 2, 3. This rule will be found useful in giving the sizes for "Unit" girders, or girders for a load = 1.0 with a length 1.0, from which the sizes necessary for any load and length are easily obtained from direct experiment on beams of any size or form (485). (463.) 8th. The breaking weights of similar cast-iron girders are to each other as the areas of their bottom flanges. Thus, in the girders D, E, F, Fig. 153, the areas of the bottom flanges are 4, 16, and 36 square inches respectively, or in the ratio 1, 4, 9, which is also the ratio of the breaking weights. (464.) 9th. The breaking weights of girders of similar sections are to each other as the areas of their bottom flanges, multiplied by the respective depths, and by a constant adapted to the particular form of section, divided by the length between supports, or (465.) (466.) (467.) In which a W = a x d x M÷l. a = (W × 1) ÷ (d × M). M = (W × 1) ÷ (a × d). the area of the bottom flange in square inches, and the rest as in (458). The value of M will vary with the form of section; for the section given by Fig. 153 it is 22.68 for tons-thus with F we have (54·441 × 180) ÷ (36 × 12) = 22.68 M. = For the form of section recommended by Mr. Hodgkinson (351) and Fig. 79, with top and bottom flanges having areas in the ratio 1 to 6, he gives the value of M at 26, for finding the breaking weight in tons. With that value for M, we obtain for the girder D, 4 × 4 × 26 ÷ 60 = 6.933 tons; for E, 16 x 8 × 26 120 = 27.73 tons; and for F, 36 × 12 × 26 180 = 62.4 tons, and those breaking weights are in the ratio 1, 4, 9, as in Fig. 153. (468.) 10th. In similar beams the weights of the beams between supports are proportional to the cubes of the linear ratio of the dimensions. Thus, in Figs. 152, 153, the linear ratio being 1, 2, 3, we have 13, 23, 33, or 1, 8, 27, which is also the ratio of the weights of the beams. We may now give some illustrations of the application of these laws to cases in practice. (469.) The 6th law in (461) states that the breaking weights of similar beams are to each other as the squares of their linear dimensions. Mr. Hodgkinson made some experiments with the special object of testing the accuracy of this law as applied to tubular beams. These experiments had particular reference to the great Britannia and Conway bridges, in which tubular beams were used for the first time on a large scale. Little or nothing was at that time known of the strength of beams of the tubular form, and it was highly necessary that the theoretical laws should be checked by experiment. In Table 78 the beams are grouped together in pairs, selected so as to be “similar" in all respects as nearly as possible, but also very different in dimensions, the special object being to show how far the strength of a large Tubular beam could be calculated with accuracy from that of its small fellow. Of course it was impossible in practice to preserve precisely any given ratio between all the dimensions; for instance, in Nos. 9 and 10 the ratio intended was 1 to 12, as in col. 6, but the actual ratios of the depths, breadths, and thicknesses of plate were 11.92, 12.31, and 12.29 respectively, as shown by cols. 7, 8, and 9. The best that can be done under these circumstances is to take the mean ratio of the four dimensions as the general ratio or value of R, and this is given by col. 10 = 12.13. (470.) Now, the experimental breaking weight of the small beam of this pair, as given by col. 2, is 1.127 ton, and 12.132 being 147.1, we obtain 1.127 × 147·1 = 165.8 tons as the strength of the large beam, col. 14. But experiment gave 118 02 tons only, hence we have 165·8 118·02 = 1·405, or an error of +40.5 per cent., as in col. 15. Calculating in this way throughout, the mean error of the whole series of experiments is + 17.58 per cent., col. 15, showing that the theoretical ratio R is too high, and we have to find the power of R which will agree with the experiments. Thus, with Nos. 9 and 10, the logarithm of 118.02 2.071889; of 1∙127 = 0·051924; and of 12·13 = 1·083861. Using these numbers logarithmically we obtain (2.071889-0·051924) 1·083861=1·864, = as in col. 11, which is the power of the ratio of the linear dimensions, or R, agreeing with the experimental strengths of this pair of beams, so that instead of R' as due by theory, we have R1-864 due to experiment, as given in col. 11, which gives throughout the power of R for each pair of beams, the mean of the whole being 1·9062. (471.) We may therefore admit that the mean power of the ratio governing the strength of "similar" tubular beams of wrought plate-iron is R instead of R2 due by theory: the effect of this apparently small difference is in extreme cases very great, as shown in (482). = = Col. 12 has been calculated by that ratio throughout; thus, in Nos. 9, 10, the mean ratio R 12.13, the logarithm of which, or 1.083861 × 1.9 = 2.059336, the natural number due to which is 114 65:-then the breaking weight of the small tube being 1.127 ton, we obtain 1·127 × 114.65 129.2 tons, the breaking weight of the large tube, as in col. 12. Experiment gave 118 02 tons, hence we have 129.2118.02 or an error of +9.5 per cent., col. 13: the mean error of the whole series as thus calculated is 0.7 per cent. only. 1.095, (472.) With the modification of R19 instead of R2, the correctness of the theoretical law is fairly borne out by the experiments. The rule was thus tested because there was some doubt whether it would hold in the case of tubular beams made with thin plates of wrought iron, which have a tendency to fail by wrinkling with a strain much lower than is necessary to crush the material. But it is shown in (316) that when the thickness and breadth of the plate subjected to compression are both increased in the same ratio, which of course is the case with "similar" beams, the Wrinkling strain per square inch remains constant, the rule (308) being Ww twbw x 104. Thus, with three plates of the respective thicknesses o, ro, 1 inch, and breadths 10, 20, 30 inches, we have 110 x 104 = × = 10.4; √220 × 104 10·4; and 330 x 104 = 10.4 tons per square inch, Ww being constant. Table 78 shows that this law is correct as applied to tubular beams, even with such a ratio in the sizes as 12 to 1, and as illustrated by Fig. 100. (473.) "Conway Tube."-We may now apply this law to the calculation of the strength of the great tube of the Conway Bridge. A large model tube was made, 75 feet long between bearings, and when the best proportions of the areas at the top and bottom had been decided by many experiments, the tube was finally broken with 86 25 tons in the centre; but the weight of the tube itself was 5.8 tons, which, being a distributed load, is equivalent to 5.822.9 tons in the centre; the tolal breaking load was therefore 86.25 +2.9 = 89.15 tons. = The breaking weight of similar beams being by the modified rule (471) proportional to R, the length of the model tube 75 feet, and of the Conway tube 400 feet, we get 89.15 × 40019 ÷7519 = 2145 tons, the details of the calculation being as follows. The logarithm of 400, or 2 602 × 1.9 4.9438, the natural number due to which, or 87,870, is the 1.9 power of 400: then the log. of 75, or 1.875 × 1·9 3.5625, the natural number due to which, or 3652, is the 1.9 power of 75. The breaking weight of the model tube, or 89.15 × 878703652 = 2145 tons in the centre, the total gross breaking weight of the Conway tube. = (474.) By the 10th law (468) the weights of similar beams are proportional to the cubes of the linear dimensions:-in our case we have 5.8 x 400 ÷ 753 880 tons: the actual weight of the Conway tube, 424 feet long, was 1146 tons, but between bearings 400 feet apart, it would be about 1146 × 400 = 424 - 1080 tons, which, being a distributed load, is equivalent to 540 tons in the centre, hence the useful load would be 2145 540 = 1605 tons in the centre. It was found necessary, in order to resist the shearing strain (403) at the bearings, to introduce about 300 tons of cast-iron frames, but as these were principally at the ends they would not sensibly increase the load upon the beam. (475.) The model tube was in this case made of the largest |